Find the distance of (-1,2,5) from the plane passing through the point (3,4,5)and whose direction-ratios are `lt 2, -3, 6 gt `.

Answer

To find the distance of the point (-1, 2, 5) from the plane passing through the point (3, 4, 5) with direction ratios (2, -3, 6), we can follow these steps:

### Step 1: Write the equation of the plane

The general equation of a plane can be expressed in the form:

\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]

where \((x_0, y_0, z_0)\) is a point on the plane and \((a, b, c)\) are the direction ratios of the normal to the plane.

Here, the point on the plane is \((3, 4, 5)\) and the direction ratios are \((2, -3, 6)\).

Substituting these values into the equation gives:

\[ 2(x - 3) - 3(y - 4) + 6(z - 5) = 0 \]

### Step 2: Simplify the equation of the plane

Expanding the equation:

\[ 2x - 6 - 3y + 12 + 6z - 30 = 0 \]

Combining like terms:

\[ 2x - 3y + 6z - 24 = 0 \]

Thus, the equation of the plane is:

\[ 2x - 3y + 6z - 24 = 0 \]

### Step 3: Use the distance formula from a point to a plane

The distance \(D\) from a point \((x_1, y_1, z_1)\) to the plane \(Ax + By + Cz + D = 0\) is given by the formula:

\[ D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

For our plane \(2x - 3y + 6z - 24 = 0\), we have:

- \(A = 2\)
- \(B = -3\)
- \(C = 6\)
- \(D = -24\)

And the point is \((-1, 2, 5)\).

### Step 4: Substitute the point into the distance formula

Substituting the point into the formula:

\[ D = \frac{|2(-1) - 3(2) + 6(5) - 24|}{\sqrt{2^2 + (-3)^2 + 6^2}} \]

Calculating the numerator:

\[ = | -2 - 6 + 30 - 24 | \]
\[ = | -2 - 6 + 30 - 24 | = | -2 - 6 + 6 | = | -2 | \]
\[ = 2 \]

Calculating the denominator:

\[ \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

### Step 5: Calculate the distance

Thus, the distance \(D\) is:

\[ D = \frac{2}{7} \]

### Final Answer

The distance of the point (-1, 2, 5) from the plane is:

\[ \frac{2}{7} \text{ units} \]